A wire of length 314 cm carrying current of 14 A is bent to form a circle. The magnetic moment of the coil is A- m 2. [Given π=3.14]

Question

A wire of length 314 cm carrying current of 14 A is bent to form a circle. The magnetic moment of the coil is A- m 2. [Given π=3.14] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7c98bad2b92fb27667ebb2f686d72c4c-.png" /> (314/100)=2 π R R =0.5 m Magnetic Moment = IA =14 × π R 2 =14 ×(3.14) × (1/4) =10.99 ≈ 11.00

Q. A wire of length $314 c m$ carrying current of $14 A$ is bent to form a circle. The magnetic moment of the coil is ____A- $m^{2}$ . [Given $π = 3.14]$

$\frac{314}{100} = 2 π RR = 0.5 m$
Magnetic Moment $= I A$
$= 14 \times π R^{2}$
$= 14 \times (3.14) \times \frac{1}{4}$
$= 10.99 \approx 11.00$

Q. A wire of length 314cm carrying current of 14A is bent to form a circle. The magnetic moment of the coil is ____A- m2. [Given π=3.14]

100314​=2πRR=0.5m Magnetic Moment =IA =14×πR2 =14×(3.14)×41​ =10.99≈11.00

Q. A wire of length $314 c m$ carrying current of $14 A$ is bent to form a circle. The magnetic moment of the coil is ____A- $m^{2}$ . [Given $π = 3.14]$

$\frac{314}{100} = 2 π RR = 0.5 m$
Magnetic Moment $= I A$
$= 14 \times π R^{2}$
$= 14 \times (3.14) \times \frac{1}{4}$
$= 10.99 \approx 11.00$